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A random sample of 20 acres gave a mean yield of wheat equal to \(41.2\) bushels per acre with a standard deviation of 3 bushels. Assuming that the yield of wheat per acre is normally distributed, construct a \(90 \%\) confidence interval for the population mean \(\mu .\)

Short Answer

Expert verified
The 90% confidence interval for the population mean yield of wheat per acre is (39.9, 42.5) bushels per acre.

Step by step solution

01

Determine the z score

The first step is to find the z score that corresponds to a 90% confidence interval. By checking the z table or using statistical software, we find that the z score for a two-sided 90% confidence interval is 1.645.
02

Calculation of the standard error

Next, calculate the standard error using the formula \(SE = \frac{s}{\sqrt{n}}\), where s is the standard deviation and n is the sample size. Substitute 3 for s and 20 for n into the formula. Compute to get SE = \(\frac{3}{\sqrt{20}}\) = 0.67.
03

Calculate the confidence interval

Finally, we can calculate the confidence interval using the formula \(\bar{x} \pm z*SE\). Substituting the values \(\bar{x} = 41.2\), z = 1.645 and SE = 0.67, we compute the interval \(41.2 \pm 1.645 * 0.67\). This gives us a lower limit of 39.9 and an upper limit of 42.5. Therefore, the 90% confidence interval for the population mean is (39.9, 42.5).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Z-score
The concept of a Z-score plays a fundamental role in statistics, especially in situations where we deal with confidence intervals, such as the one in the exercise provided. A Z-score is a measure that describes how many standard deviations a data point is from the mean. It's a way of standardizing scores across different data sets, which allows us to estimate probabilities, make decisions, and create predictions.
A 90% confidence interval suggests that we're looking for values that exist in a range where the true population mean is likely to fall 90% of the time. For a two-sided confidence interval like this one, the Z-score values that border 5% on either side can be found in a Z-table, or calculated using statistical software. For our exercise, this Z-score is 1.645.
Remember:
  • This value of 1.645 means that we are cutting off the outer 5% (in total) of a standard normal distribution.
  • The Z-score bridges the sample data to the normal distribution, helping us understand the dispersion regarding the mean.
Standard Error
Standard Error (SE) is an essential concept in creating a confidence interval as it quantifies the amount by which the sample mean, derived from a dataset, is expected to differ from the true population mean. In simple terms, it tells us how much sampling variability is expected.
The larger the sample size, the smaller the standard error, because having more data points usually gives us a better estimate of the population mean. The formula used in the exercise to calculate SE is: \[ SE = \frac{s}{\sqrt{n}} \]where \(s\) is the sample standard deviation and \(n\) is the sample size. Substituting the given values, we find that SE equals 0.67 for our data set.
This low standard error indicates that the sample mean is reasonably close to the population mean, making it reliable for making predictions about the entire population.
Population Mean
The Population Mean, often represented by the symbol \(\mu\), is the average of a set of characteristics for an entire population. It's a central concept in statistics, providing a single value summary that best represents a data set.
In the exercise context, you don't know the actual value of the population mean. That's where confidence intervals help us estimate it based on a sample. With the sample mean at 41.2 bushels per acre given in the problem, we create an interval that the population mean is likely to fall into, thus helping us understand the overall tendency of the data.
An accurate estimation of the population mean is crucial, especially in agriculture where understanding crop yield can influence everything from planning and resource allocation to financial decisions and food security.
Normal Distribution
Normal Distribution, also referred to as the bell curve due to its shape, is one of the most important probability distributions in statistics. It describes how values of a variable are likely to be distributed, showing a symmetric distribution where most observations cluster around a central peak, with probabilities tapering off equally towards both ends.
The exercise assumes a normal distribution of wheat yield, which is vital because the calculation of a confidence interval relies heavily on the assumption that the data is normally distributed.
Here's what makes Normal Distribution significant for confidence intervals:
  • It allows us to apply the Z-score, which assumes normality, to determine the likelihood of data points appearing within specific bounds.
  • The standard properties dictate that about 68% of values fall within one standard deviation of the mean, about 95% fall within two, and about 99.7% fall within three. This helps us gauge the spread and central tendency of data.
Understanding normal distribution assists not only in calculating confidence intervals but also provides broader insights into data variability and prediction modeling.

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Most popular questions from this chapter

You are interested in estimating the mean commuting time from home to school for all commuter students at your school. Briefly explain the procedure you will follow to conduct this study. Collect the required data from a sample of 30 or more such students and then estimate the population mean at a \(99 \%\) confidence level. Assume that the population standard deviation for such times is \(5.5\) minutes.

An article in the Los Angeles Times (latimesblogs.latimes.com/pardonourdust/) quoted from the National Association of Realtors that "we now sell our homes and move an average of every six years." Suppose that the average time spent living in a house prior to selling it for a random sample of 400 recent home sellers was \(6.18\) years and the sample standard deviation was \(2.87\) years. a. What is the point estimate of the corresponding population mean? b. Construct a \(98 \%\) confidence interval for the average time spent living in a house prior to selling it for all home owners. What is the margin of error for this estimate?

a. A sample of 1100 observations taken from a population produced a sample proportion of \(.32 .\) Make a \(90 \%\) confidence interval for \(p\). b. Another sample of 1100 observations taken from the same population produced a sample proportion of .36. Make a \(90 \%\) confidence interval for \(p\). c. A third sample of 1100 observations taken from the same population produced a sample proportion of .30. Make a \(90 \%\) confidence interval for \(p\). d. The true population proportion for this population is \(.34 .\) Which of the confidence intervals constructed in parts a through c cover this population proportion and which do not?

a. How large a sample should be selected so that the margin of error of estimate for a \(99 \%\) confidence interval for \(p\) is \(.035\) when the value of the sample proportion obtained from a preliminary sample is \(.29\) ? b. Find the most conservative sample size that will produce the margin of error for a \(99 \%\) confidence interval for \(p\) equal to \(.035\).

A Centers for Disease Control and Prevention survey about cell phone use noted that \(14.7 \%\) of U.S. households are wireless-only, which means that the household members use only cell phones and do not have a landline. Suppose that this percentage is based on a random sample of 855 U.S. households (Source: http://www.cdc.gov/nchs/data/nhsr/nhsr014.pdf). a. Construct a \(95 \%\) confidence interval for the proportion of all U.S. households that are wireless-only. b. Explain why we need to construct a confidence interval. Why can we not simply say that \(14.7 \%\) of all U.S. households are wireless-only?

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